Magnetic moment

Can someone prove that for a hollow sphere given a uniformly distributed charge Q and rotated with ang. vel. ω the magnetic moment of sphere =
(Q/3)* r2ω where r is radius

10 Answers

11
rkrish ·

Angular Momentum = L (let) = (2/3) mr2. ω
Magnetic Monent = M (let)

For any regular body with uniform mass & charge distribution rotating with ω :

M/L = Q/(2m) m : mass of body.

So,for this case,

M = Q((2/3)mr2. ω) / (2m) = (Q/3) . r2 ω

33
Abhishek Priyam ·

u can do this by integrations... by assuming rings...

or M/L=q/2m

M is mag mom... L is ang momentum about COM... q is charge.. m is mass...

33
Abhishek Priyam ·

din saw ur reply :P

1
Philip Calvert ·

@ rkrish
arre that's why i am asking the question
and u are underlining i don't know what
i asked for proof and this is not it

give me a proof for hollow sphere

@ abhi can u give me the proof
i have been tru all this b4 and i am asking bcozt i had problem with integration of those "rings" why don't u help me with that part

33
Abhishek Priyam ·

dekhte hain...

1
Philip Calvert ·

anyone ....
i have problems in integrating this...
ok can someone prove the thing for any rigid body in general ??(if it is true for all that is)

33
Abhishek Priyam ·

Charge on ring.. is Qsinθdθ
2

i=Qsinθdθω/2π
2

idA=Qsinθdθ ωπ(Rsinθ)2
2 2Ï€

M=∫idA

=QR2ω 2 0∫π/2sin3θdθ
4
QR2ω/3

33
Abhishek Priyam ·

philip...

is it ok

1
Philip Calvert ·

ok thanks i tried it three times but with no result somehow i was getting sin^4 there..
it must be ok .. anyway it was just curiosity u know..

33
Abhishek Priyam ·

:)

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